# A line segment is bisected by a line with the equation  8 y + 5 x = 4 . If one end of the line segment is at ( 2 , 7 ), where is the other end?

Sep 23, 2017

$\left(- \frac{442}{89} , - \frac{369}{89}\right)$

#### Explanation: Let $L 1$ be the perpendicular bisector and $L 2$ be the bisected line, as shown in the figure.
Given that the equation of the bisector $L 1$ is $8 y + 5 x = 4$,
$\implies y = - \frac{5}{8} x + \frac{1}{2}$
Let ${m}_{1}$ be the slope of $L 1$, and ${m}_{2}$ the slope of $L 2$,
$\implies {m}_{1} = - \frac{5}{8}$
Recall that the product of the slopes of two perpendicular lines is $- 1$,
$\implies {m}_{2} \times {m}_{1} = - 1 , \implies {m}_{2} = \frac{8}{5}$
$\implies$ equation of $L 2$ is : $y - 7 = \frac{8}{5} \left(x - 2\right)$
$\implies y = \frac{8}{5} x + \frac{19}{5}$
Set the equations of $L 1 \mathmr{and} L 2$ equal to each other to find the intersection point $P \left({x}_{m} , {y}_{m}\right)$, which is also the midpoint of $L 2$.
$\implies - \frac{5}{8} x + \frac{1}{2} = \frac{8}{5} x + \frac{19}{5}$
$\implies x = - \frac{132}{89}$
$\implies y = - \frac{5}{8} x + \frac{1}{2} = - \frac{5}{8} \times \left(- \frac{132}{89}\right) + \frac{1}{2} = \frac{127}{89}$
$\implies P \left({x}_{m} , {y}_{m}\right) = \left(- \frac{132}{89} , \frac{127}{89}\right)$
Let the other end point of $L 2$ be $B \left(x , y\right)$,
Since $P$ is the midpoint of $L 2$,
$\implies \left(\frac{x + 2}{2} , \frac{y + 7}{2}\right) = \left(- \frac{132}{89} , \frac{127}{89}\right)$
$\implies \left(x , y\right) = \left(- \frac{442}{89} , - \frac{369}{89}\right)$